If int e^{2x} frac{2(sin 2x cos 2x - 1)}{2 si | vedclass

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(A) The integral is $I = \int e^{2x} \frac{2(\sin 2x \cos 2x - 1)}{2 \sin^2 2x} dx = \int e^{2x} \frac{\sin 2x \cos 2x - 1}{\sin^2 2x} dx$.$I = \int e

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$1/8$

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(A) The integral is $I = \int e^{2x} \frac{2(\sin 2x \cos 2x - 1)}{2 \sin^2 2x} dx = \int e^{2x} \frac{\sin 2x \cos 2x - 1}{\sin^2 2x} dx$.$I = \int e^{2x} (\frac{\sin 2x \cos 2x}{\sin^2 2x} - \frac{1}{\sin^2 2x}) dx = \int e^{2x} (\cot 2x - \csc^2 2x) dx$.Let $f(x) = \cot 2x$. Then $f'(x) = -\csc^2 2x \cdot 2 = -2 \csc^2 2x$.This does not fit the form $\int e^{ax}(f(x) + f'(x)/a) dx$. Let's rewrite: $I = \int e^{2x} \cot 2x dx - \int e^{2x} \csc^2 2x dx$.Using integration by parts on the first term: $\int e^{2x} \cot 2x dx = \cot 2x \cdot \frac{e^{2x}}{2} - \int (-\csc^2 2x \cdot 2) \cdot \frac{e^{2x}}{2} dx = \frac{1}{2} e^{2x} \cot 2x + \int e^{2x} \csc^2 2x dx$.Substituting this back: $I = (\frac{1}{2} e^{2x} \cot 2x + \int e^{2x} \csc^2 2x dx) - \int e^{2x} \csc^2 2x dx = \frac{1}{2} e^{2x} \cot 2x + c$.Comparing with $A e^{2x} \cot 2x + c$, we get $A = 1/2$.Therefore, $A^3 = (1/2)^3 = 1/8$.

If $\int e^{2x} \frac{2(\sin 2x \cos 2x - 1)}{2 \sin^2 2x} dx = A e^{2x} \cot 2x + c$ (where $c$ is the constant of integration), then $A^3 =$ ?

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